A plane P is parallel to two lines whose direction ratios are $$-2,1,-3$$ and $$-1,2,-2$$ and it contains the point $$(2,2,-2)$$. Let P intersect the co-ordinate axes at the points $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ making the intercepts $$\alpha, \beta, \gamma$$. If $$\mathrm{V}$$ is the volume of the tetrahedron $$\mathrm{OABC}$$, where $$\mathrm{O}$$ is the origin, and $$\mathrm{p}=\alpha+\beta+\gamma$$, then the ordered pair $$(\mathrm{V}, \mathrm{p})$$ is equal to :

The foot of the perpendicular from a point on the circle $$x^{2}+y^{2}=1, z=0$$ to the plane $$2 x+3 y+z=6$$ lies on which one of the following curves?

If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$$ is $$2 \sqrt{6}$$ units and $$Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$$ is the image of the point P in this line, then $$\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$$ is equal to :

If the line of intersection of the planes $$a x+b y=3$$ and $$a x+b y+c z=0$$, a $$>0$$ makes an angle $$30^{\circ}$$ with the plane $$y-z+2=0$$, then the direction cosines of the line are :